If q cos θ = p, find tan θ - cot θ in terms of p and q.

Let ABC be a right angle triangle with ∠B = 90° and ∠C = θ.

Given,

⇒ q cos θ = p

⇒ cos θ = $\dfrac{p}{q}$ ……….(1)

By formula,

⇒ cos θ = $\dfrac{\text{Base}}{\text{Hypotenuse}}$

⇒ cos θ = $\dfrac{BC}{AC}$ ……….(2)

Comparing equations (1) and (2) we get :

$\Rightarrow \dfrac{p}{q} = \dfrac{BC}{AC}$

Let BC = pk and AC = qk.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ (qk)² = AB² + (pk)²

⇒ AB² = q²k² - p²k²

⇒ AB² = k²(q² - p²)

⇒ AB = $\sqrt{k^2(q^2 - p^2)} = k\sqrt{q^2 - p^2}$.

By formula,

$\text{tan θ} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{AB}{BC} = \dfrac{k\sqrt{q^2 - p^2}}{pk} \\[1em] = \dfrac{\sqrt{q^2 - p^2}}{p}. \\[1em] \text{cot θ} = \dfrac{1}{\text{tan θ}} \\[1em] = \dfrac{1}{\dfrac{\sqrt{q^2 - p^2}}{p}} \\[1em] = \dfrac{p}{\sqrt{q^2 - p^2}}.$

Substituting values in tan θ - cot θ we get :

$\Rightarrow \text{tan θ - cot θ} = \dfrac{\sqrt{q^2 - p^2}}{p} - \dfrac{p}{\sqrt{q^2 - p^2}} \\[1em] = \dfrac{\sqrt{q^2 - p^2}\sqrt{q^2 - p^2} - p^2}{p\sqrt{q^2 - p^2}} \\[1em] = \dfrac{q^2 - p^2 - p^2}{p\sqrt{q^2 - p^2}} \\[1em] = \dfrac{q^2 - 2p^2}{p\sqrt{q^2 - p^2}}.$

Hence, tan θ - cot θ = $\dfrac{q^2 - 2p^2}{p\sqrt{q^2 - p^2}}.$